Question

Find the equation of the tangent to the graph at the indicated point. HINT [Compute the derivative algebraically]
f(x) = x^2 − 1; a = 1


I found the slope using lim as h approaches 0, f(1+h) - f(1) / h and got lost during that part.

Answer 1

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h)- f(x)}{h}\]

for x = 1:

\[
\lim_{h \to 0} \frac{f(1+h)- f(1)}{h}
\lim_{h \to 0} \frac{(1+h)^2 -1 - (1^2 - 1)}{h} =
\lim_{h \to 0} \frac{1+2h + h^2 -1 - 1 + 1)}{h} =
\]

\[
\lim_{h \to 0} \frac{2h+h^2}{h} = \lim_{h \to 0} \frac{h(2+h)}{h} = \lim_{h \to 0} \,\,(2+h) = 2
\]

Answer 2

\[ y-y_1 = m(x - x_1) \]

\[ y_1 = f(1) = 1^2 - 1= 0\]
equation of the tangent line at that point therefore is

\[ y = 2(x-1) \]
\[ y = 2x-2 \]

Answer 3

That's what I ended up getting for the second part. Thank you!

Answer 4

you're welcome